= Solution
Fix a bounded tile in <Euclidean space> $\Omega$ with labelled flat boundary faces. For each face label $s$, encode an assembly of $n$ copies by an involutive real symmetric <signed permutation matrix> $M_s$: a paired interface contributes a two-by-two swap, an unglued face with a <Neumann boundary condition> contributes $+1$ on the diagonal, and an unglued face with a <Dirichlet boundary condition> contributes $-1$. Use the same face-identification convention on each copy. If a second assembly has matrices $N_s$ and there is an invertible constant matrix $C$ satisfying
$$
\boxed{CM_s=N_sC\quad\hbox{for every face label }s,}
$$
then the map $u=(u_1,\ldots,u_n)\mapsto Cu$ is an isomorphism between corresponding <Laplace-Beltrami operator> <eigenspaces>. This is the <transplantation theorem>.
To prove it, restrict an <eigenfunction> to the tile copies and identify each copy with $\Omega$. Along a face, let $a$ be its vector of boundary values and $b$ its vector of outward <normal derivatives>. At a glued interface the values agree and the outward derivatives are opposite. At an unglued face with a <Neumann boundary condition> the derivative vanishes, and at a face with a <Dirichlet boundary condition> the value vanishes. All three cases are encoded by
$$
M_sa=a,\qquad M_sb=-b.
$$
The intertwining identities give $N_s(Ca)=Ca$ and $N_s(Cb)=-Cb$, so the transplanted function satisfies the second assembly's interface and <boundary conditions>. On each tile its <Laplace-Beltrami operator> is the same constant linear combination of the original Laplacians, hence it has the same <eigenvalue>. Across an interface, matching values and opposite <normal derivatives> make it a weak solution; <elliptic regularity>, or the allowed reflection principle, makes it a genuine smooth solution there. At corners the operator is understood by its finite-energy, <Friedrichs extension>; the same matching argument applies to its <quadratic form> domain. Applying $C^{-1}$ reverses the construction. Thus dimensions of all <eigenspaces> agree, including <multiplicities>, and the two specified boundary-value spectra coincide.
\b[The requested pair exists with explicitly specified mixed boundary conditions and polygonal boundary.] Take four copies $Q_0,Q_1,Q_2,Q_3$ of an octagonal <regular polygon> in <Euclidean space>. Label four alternate sides $a,b,c,d$ in cyclic order; the four intervening sides always have a <Neumann boundary condition>. The labelled sides are disjoint even at their endpoints. Parameterize every side in the positive boundary direction of the reference octagon. When pairing same-labelled sides, identify equal parameters. This produces a locally flat <surface with boundary>: a neighbourhood of an interior point of an interface consists of two half-disks glued along their diameters. All tile vertices remain boundary points, with boundary angles $3\pi/4$ or $3\pi/2$; there are no interior cone points.
Here are the complete gluing and <boundary conditions>. An entry $ij$ pairs the corresponding sides of $Q_i,Q_j$. The indicated unpaired sides have a <Dirichlet boundary condition>; every other unpaired side has a <Neumann boundary condition>.
$$
\begin{array}{c|c|c|c|c}
\text{face}&\text{pairs in }D&\text{Dirichlet tiles in }D&\text{pairs in }B&\text{Dirichlet tiles in }B\\ \hline
a&23&\varnothing&13&\varnothing\\
b&12&\varnothing&12&\varnothing\\
c&01&\{2,3\}&02&\{1,3\}\\
d&\varnothing&\{1,3\}&01,\ 23&\varnothing
\end{array}
$$
Both assemblies are compact and connected. The <Laplace-Beltrami operator> is the nonnegative operator $P=-\Delta$ associated with the finite-energy <quadratic form> $\int|\nabla u|^2$, with zero <Sobolev trace> on the specified Dirichlet arcs. Thus the <mixed boundary condition> is defined without requiring classical differentiability at a corner or at a Dirichlet–Neumann junction.
The four <signed permutation matrices> for $D$ are
$$
M_a=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0\end{pmatrix},\quad
M_b=\begin{pmatrix}1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{pmatrix},\quad
M_c=\begin{pmatrix}0&1&0&0\\1&0&0&0\\0&0&-1&0\\0&0&0&-1\end{pmatrix},\quad
M_d=\operatorname{diag}(1,-1,1,-1).
$$
For $B$ they are
$$
N_a=\begin{pmatrix}1&0&0&0\\0&0&0&1\\0&0&1&0\\0&1&0&0\end{pmatrix},\quad
N_b=M_b,\quad
N_c=\begin{pmatrix}0&0&1&0\\0&-1&0&0\\1&0&0&0\\0&0&0&-1\end{pmatrix},\quad
N_d=\begin{pmatrix}0&1&0&0\\1&0&0&0\\0&0&0&1\\0&0&1&0\end{pmatrix}.
$$
Use the <Hadamard matrix>
$$
\boxed{C=\begin{pmatrix}1&1&1&1\\1&-1&1&-1\\1&1&-1&-1\\1&-1&-1&1\end{pmatrix},\qquad C^{-1}=\frac14C.}
$$
The intertwining relations can be checked on all columns at once. Writing the columns of $C$ as $h_0,h_1,h_2,h_3$, direct application of the displayed $N_s$ gives
$$
\begin{aligned}
N_aC&=(h_0,h_1,h_3,h_2)=CM_a,\\
N_bC&=(h_0,h_2,h_1,h_3)=CM_b,\\
N_cC&=(h_1,h_0,-h_2,-h_3)=CM_c,\\
N_dC&=(h_0,-h_1,h_2,-h_3)=CM_d.
\end{aligned}
$$
For the four intervening faces both assemblies have the identity face matrix, so these intertwining relations hold as well. Also $C^TC=C^2=4I$. This <four-tile mixed-boundary transplantation between a disk and a nonorientable surface> gives a bijection of every pair of <eigenspaces>, preserving <eigenvalues> and <multiplicities>.
There is also a direct argument on the <quadratic form> domains that handles all boundary corners. The <orthogonal matrix> $U=C/2$ acts on tile restrictions, preserves their <L2 inner product> and the sum of their <Dirichlet energies>, and commutes with taking a <Sobolev trace>. The condition $M_su=u$ on each face encodes equality of the two traces at a paired interface and vanishing trace at an unpaired Dirichlet face. Intertwining takes these conditions exactly to $N_s(Uu)=Uu$. Consequently $U$ maps the whole form domain onto the other form domain and preserves its <quadratic form>. The associated <self-adjoint operators> are unitarily equivalent, proving isospectrality for the stated <mixed boundary conditions> without any assumption of smoothness at a Dirichlet–Neumann junction.
We now verify the actual topology using the <topology of disk tiles glued along disjoint boundary arcs>. Because the gluing arcs in each tile are disjoint, the resulting <surface with boundary> is a thickening of its gluing <graph>, with one vertex for each <closed disc> tile and one edge for each paired arc. One can construct this <homotopy equivalence> by taking a small central <closed disc> in each tile and narrow strips to its paired arcs. What remains is a <collar neighbourhood> of the boundary, which retracts onto this connected system of <closed discs> and strips. Shrinking each central <closed disc> and each strip then gives the gluing <graph>. Equivalently, begin with a <spanning tree> of the gluing <graph>: its <closed disc> tiles joined along boundary arcs form one <closed disc>, and every remaining pair of boundary arcs adds one band. Thus for $n$ tiles and $e$ paired arcs the <fundamental group> is a <free group> of rank $e-n+1$, and the <Euler characteristic> is $n-e$.
For $D$, the gluing <graph> is the path
$$
0\xrightarrow{c}1\xrightarrow{b}2\xrightarrow{a}3.
$$
Gluing a new <closed disc> to a <closed disc> along one boundary arc again gives a <closed disc>; applying this three times proves that $D$ is homeomorphic to a <closed disc>. In particular
$$
\boxed{D\text{ is orientable and simply connected},\qquad\chi(D)=1,\qquad\pi_1(D)=\{1\}.}
$$
For $B$, the five paired arcs give the <graph> edges $13,12,02,01,23$. It is connected, and hence
$$
\boxed{\chi(B)=4-5=-1,\qquad\pi_1(B)\cong F_2.}
$$
Here $F_2$ is the <free group> on two generators, so $B$ is not <simply connected>. To decide <orientation of a surface>, give every reference tile its counterclockwise <orientation>. Our same-parameter gluing preserves boundary directions, so a global <orientation> requires opposite orientation signs on paired tiles: the two induced boundary orientations at an interior interface must be opposite. Such signs exist exactly when the gluing <graph> is <bipartite>. In $B$ the cycle
$$
1\xrightarrow{b}2\xrightarrow{d}3\xrightarrow{a}1
$$
has length three. Following it forces the sign to reverse three times and return to its starting tile with the opposite sign. This proves that $B$ is a <nonorientable surface>. It has nonempty boundary because every intervening octagon side remains unpaired. Thus \b[$D$ is a simply-connected orientable surface with boundary, whereas its isospectral partner $B$ is neither simply connected nor orientable], exactly as required, for the specified mixed boundary problem.
The boundary qualification is essential. With smooth boundary and the same pure <Dirichlet boundary condition> or pure <Neumann boundary condition> on both surfaces, the following existing obstruction applies. Polygonal corners and the <mixed boundary conditions> in the construction above place it outside that obstruction.
With the interpretation of compact smooth <Riemannian surfaces> with the same pure <Dirichlet boundary condition> or pure <Neumann boundary condition>, the printed simply-connected/nonorientable pair cannot exist. The two-dimensional <heat trace> has the following expansion (the smooth-boundary coefficients are given in https://pages.uoregon.edu/gilkey/dirPDF/E27Handbook.pdf[the heat-coefficient formulas, Theorems 5.1–5.2]):
$$
\operatorname{Tr}(e^{-tP})
=\frac{A}{4\pi t}\mp\frac{L}{8\sqrt{\pi t}}
+\frac1{12\pi}\left(\int_M K\,dA+\int_{\partial M}\kappa\,ds\right)+O(t^{1/2}),
$$
where the boundary sign is minus for Dirichlet and plus for Neumann. The <Gauss-Bonnet theorem> identifies the constant term with $\chi(M)/6$. Isospectrality therefore forces equal <Euler characteristics>. A compact connected simply-connected surface with boundary is a disk and has $\chi=1$. A connected nonorientable surface with $c\geq1$ crosscaps and $b\geq1$ boundary components has $\chi=2-c-b\leq0$. Thus \b[no such smooth pure-Dirichlet or pure-Neumann pair exists]. This is the <Euler characteristic from the heat trace of a bordered surface>.
Piecewise-smooth boundaries introduce corner terms, and mixed <boundary conditions> change the heat coefficients, as the explicit construction above demonstrates. Moreover, in the usual pure-Neumann polygonal reflection-transplantation setup, the <pure Neumann reflection transplantation preserves Euler characteristic> argument applies even with corners. The numbers of tiles agree. For each face label the number of edge classes is $(n+\operatorname{tr}M_s)/2$, so these agree too. For each tile-vertex label, vertex classes are orbits of the <subgroup> generated by incident-face involutions. Their number is the <dimension> of invariant vectors in that permutation representation, which is unchanged by the intertwiner. Hence the numbers of vertices agree, proving equal $V-E+F$. Thus this standard tiled construction also cannot supply a disk on one side and a nonorientable surface on the other.
For comparison, the following existing construction gives an <orientable> and a nonorientable pair with the same <spectrum> for the pure <Neumann boundary condition>. Both have nontrivial <fundamental groups>; this is a separate example, not the construction used to meet the simply-connected requirement above. This <eight-tile Neumann transplantation across orientability> makes the comparison explicit. The finite seed is the affine modulo-eight example in https://arxiv.org/pdf/2008.12498[the original orientability construction]; the intertwiner and the topology checks are derived below. Take eight copies of a regular Euclidean hexagon and label three pairwise nonadjacent sides $\sigma,t,u$. Every remaining side is a side with a <Neumann boundary condition>. Index the copies by $j\in\mathbb Z/8\mathbb Z$. For the first assembly pair same-labelled faces according to
$$
\sigma(j)=1-j,\qquad t(j)=-j,\qquad u_1(j)=3j;
$$
for the second, use
$$
\sigma(j)=1-j,\qquad t(j)=-j,\qquad u_2(j)=3j+4,
$$
with all arithmetic modulo eight. A fixed point means an unglued side with a <Neumann boundary condition>. Pair faces by the reflection convention, matching the same endpoint parameter in the reference tiles. The distinguished sides are disjoint, so no interior vertex singularities are created; the resulting surfaces have polygonal boundary corners.
Let $R e_j=e_{j+1}$ on the eight-dimensional tile-index space and set
$$
\boxed{C=I+R^4+2(R+R^{-1}).}
$$
It commutes with the <permutation matrices> for $t$ and $\sigma$, because its coefficients are invariant under reversing cyclic order. For $u_1,u_2$, write $C_{ij}=c_{i-j}$, where $c_0=c_4=1$, $c_1=c_7=2$, and all other $c_r$ vanish. The required relation $CM_{u_1}=M_{u_2}C$ is exactly $c_r=c_{3r+4}$, verified from these four nonzero entries. Thus all three intertwining relations hold.
The cyclic Fourier vectors diagonalize $C$. Its <eigenvalues> are
$$
1+(-1)^k+4\cos(\pi k/4),\qquad0\leq k\leq7,
$$
namely $6,2\sqrt2,2,-2\sqrt2,-2,-2\sqrt2,2,2\sqrt2$. None is zero, so $C$ is invertible and the two assemblies are <isospectral manifolds> for the <Neumann boundary condition>.
We prove their topology as well. Their gluing graphs are connected: the first contains the spanning path $0,1,3,6,2,7,5,4$, and the second contains the spanning cycle $0,1,7,2,6,3,5,4,0$. Each has eight tile vertices and ten paired-face edges. Since tiles are disks and gluing arcs are disjoint, each surface retracts up to <homotopy> to this multigraph; thus its <Euler characteristic> is $8-10=-2$ and its <fundamental group> is a <free group> of rank three.
An <orientation> assigns a sign to each reference tile, and a reflection gluing requires opposite signs at the ends of every graph edge. The second graph admits this assignment: take sign $+$ on $\{0,5,6,7\}$ and sign $-$ on $\{1,2,3,4\}$. Every paired face joins opposite signs, proving orientability. In the first graph the cycle
$$
1\xrightarrow{u_1}3\xrightarrow{\sigma}6\xrightarrow{t}2
\xrightarrow{\sigma}7\xrightarrow{t}1
$$
has odd length. Transporting <orientation> around it reverses the initial sign, proving nonorientability. Hence the first surface is also not <simply connected>. The second surface is <orientable> but likewise not <simply connected>; the disk required in the question is supplied instead by the four-tile mixed-boundary construction above.
The four-tile construction resolves the printed simply-connected requirement under explicit <mixed boundary conditions>. The eight-tile comparison only concerns <orientation of a surface> under a pure <Neumann boundary condition>.
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